Function is Closed? While reading the optimization textbook, the below proposition given:
Let $f: X \mapsto [-\infty, \infty]$ be a function. If $ \text{dom}(f)$ is closed and $f$ is lower semicontinuous at each x$\in \text{dom}(f)$, then $f$ is closed.
dom($f$) refers to the effective domain of f where $f(x) < \infty$ $\forall x \in$ dom($f$)
What is the closed of functions?

Miscellaneous question:
When denoting function, {domain $\to$ codomain} is different from {domain $\mapsto$ codomain}?
Because this textbook never uses the former one but later ones only.
 A: There seems to be two competing definitions of "closed" with respect to functions:  
(i) This page defines a closed map as one that maps closed sets to closed sets (but that definition would not make your theorem true, so that is likely not the correct definition in this context): 
http://mathonline.wikidot.com/open-and-closed-maps-on-topological-spaces
To see how this definition fails your theorem, consider the counter-example lower-semicontinuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ given by: 
$$ f(x)= \left\{ \begin{array}{ll}
0&\mbox{ if $x \leq 0$} \\
x+1  & \mbox{ if $x>0$} 
\end{array}
\right.$$
then consider the closed set $[0,1]$ and its mapped set $f([0,1])$. 
(ii) This page defines a closed function as one such that, for each $\alpha \in \mathbb{R}$, the set $\{x \in dom(f): f(x) \leq \alpha\}$ is closed.  This is not equivalent to the first definition, yet, it makes your theorem true.  So it must be the correct definition in this context. 
https://en.wikipedia.org/wiki/Closed_convex_function
It is remarkable (and annoying) that your textbook does not bother to define "closed function." 
